An independent-measures research study was used to compare two treatment conditions with n = 12 participants in each treatment. The first treatment had a mean of M = 55 with variance s2 = 8, and the second treatment had M = 52 and s2 = 4.

a) Use a two-tailed test with Alpha = 0.05 to determine whether these date indicate a significant difference between the two treatments.

b) Use a two-tailed test with Alpha = 0.01 to determine whether these date indicate a significant difference between the two treatments.

c) Use a one-tailed test with Alpha = 0.05 to determine whether these date indicate a significant difference between the two treatments.

Question

Mathematics

Keira Jefferson

13 December, 23:15

An independent-measures research study was used to compare two treatment conditions with n = 12 participants in each treatment. The first treatment had a mean of M = 55 with variance s2 = 8, and the second treatment had M = 52 and s2 = 4.

a) Use a two-tailed test with Alpha = 0.05 to determine whether these date indicate a significant difference between the two treatments.

b) Use a two-tailed test with Alpha = 0.01 to determine whether these date indicate a significant difference between the two treatments.

c) Use a one-tailed test with Alpha = 0.05 to determine whether these date indicate a significant difference between the two treatments.

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Answers (1)

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Myla Manning · Answer 1

(a) The data indicate a significant difference between the two treatments.

(b) The data do not indicate a significant difference between the two treatments.

(c) The data indicate a significant difference between the two treatments.

Step-by-step explanation:

Null hypothesis: There is no difference between the two treatments.

Alternate hypothesis: There is a significant difference between the two treatments.

Data given:

M1 = 55

M2 = 52

s1^2 = 8

s2^2 = 4

n1 = 12

n2 = 12

Pooled variance = [ (n1-1) s1^2 + (n2-1) s2^2] : (n1+n2-2) = [ (12-1) 8 + (12-1) 4] : (12+12-2) = 132 : 22 = 6

Test statistic (t) = (M1 - M2) : sqrt [pooled variance (1/n1 + 1/n2) ] = (55 - 52) : sqrt[6 (1/6 + 1/6) ] = 3 : 1.414 = 2.122

Degree of freedom = n1+n2-2 = 12+12-2 = 22

(a) For a two-tailed test with a 0.05 (5%) significance level and 23 degrees of freedom, the critical values are - 2.069 and 2.069.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 falls outside the region bounded by the critical values.

(b) For a two-tailed test with a 0.01 (1%) significance level and 23 degrees of freedom, the critical values are - 2.807 and 2.807.

Conclusion:

Fail to reject the null hypothesis because the test statistic 2.122 falls within the region bounded by the critical values.

(c) For a one-tailed test with 0.05 (5%) significance level and 23 degrees of freedom, the critical value is 1.714.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 is greater than the critical value 1.714.